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Home  >>  CBSE XI  >>  Math  >>  Statistics
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Find the mean deviation about the mean for the data $36,72,46,42,60,45,53,46,51,49$

$\begin{array}{1 1}(A)\;3\\(B)\;4\\(C)\;7\\(D)\;6 \end{array} $

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1 Answer

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Toolbox:
  • First arrange the given observations in ascending order and then calculate the median (n) after that mean deviation about the mean can be calculated by the reaction $ =\large\frac{\sum |x -M|}{n}$ where $n$=number of terms
Arranging the data in ascending order.
Number of observations $=10(even)$
Median $M=\Large\frac {N/2 observation + (N/2+1)th observation}{2}$
$\qquad= \Large\frac{(\large\frac{10}{2}) th \;observation + (\Large\frac{10}{2} +1)th \;observation}{2}$
$\qquad= \large\frac{5\;th\; observation +6th \;observation }{2}$
$\qquad= \large\frac{46+49}{2}$
$\qquad= 47.5 $
Step 2:
$X_i =36 , \qquad |X_i -M |=|36-47.5|=11.5$
$X_i =42 , \qquad |X_i -M |=|42-47.5|=5.5$
$X_i =45 , \qquad |X_i -M |=|45-47.5|=2.5$
$X_i =46 , \qquad |X_i -M |=|46-47.5|=1.5$
$X_i =46 , \qquad |X_i -M |=|46-47.5|=1.5$
$X_i =49 , \qquad |X_i -M |=|49-47.5|=2.5$
$X_i =51 , \qquad |X_i -M |=|51-47.5|=3.5$
$X_i =53 , \qquad |X_i -M |=|53-47.5|=5.5$
$X_i =60 , \qquad |X_i -M |=|60-47.5|=12.5$
$X_i =72 , \qquad |X_i -M |=|72-47.5|=24.5$
$\sum |x_i-M|=70$
Mean deviation about median $\large\frac{\sum|x_i-M|}{n}$
$\qquad=\large\frac{70}{10}$$=7$
Hence C is the answer.
answered Jun 25, 2014 by meena.p
 
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